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Unformatted text preview: Study Set  December 4, 2007. Problem Solutions : Yates and Goodman, 6.5.1 6.5.2 6.6.2 6.7.2 7.1.4 7.2.3 7.3.1 and 7.4.3 Problem 6.5.1 Solution (a) From Table 6.1, we see that the exponential random variable X has MGF φ X ( s ) = λ λ s (1) (b) Note that K is a geometric random variable identical to the geometric random variable X in Table 6.1 with parameter p = 1 q . From Table 6.1, we know that random variable K has MGF φ K ( s ) = (1 q ) e s 1 qe s (2) Since K is independent of each X i , V = X 1 + ··· + X K is a random sum of random variables. From Theorem 6.12, φ V ( s ) = φ K (ln φ X ( s )) = (1 q ) λ λ s 1 q λ λ s = (1 q ) λ (1 q ) λ s (3) We see that the MGF of V is that of an exponential random variable with parameter (1 q ) λ . The PDF of V is f V ( v ) = (1 q ) λe (1 q ) λv v ≥ otherwise (4) Problem 6.5.2 Solution The number N of passes thrown has the Poisson PMF and MGF P N ( n ) = (30) n e 30 /n ! n = 0 , 1 , . . . otherwise φ N ( s ) = e 30( e s 1) (1) Let X i = 1 if pass i is thrown and completed and otherwise X i = 0. The PMF and MGF of each X i is P X i ( x ) = 1 / 3 x = 0 2 / 3 x = 1 otherwise φ X i ( s ) = 1 / 3 + (2 / 3) e s (2) The number of completed passes can be written as the random sum of random variables K = X 1 + ··· + X N (3) 1 Since each X i is independent of N , we can use Theorem 6.12 to write φ K ( s ) = φ N (ln φ X ( s )) = e 30( φ X ( s ) 1) = e 30(2 / 3)( e s 1) (4) We see that K has the MGF of a Poisson random variable with mean E [ K ] = 30(2 / 3) = 20, variance Var[ K ] = 20, and PMF P K ( k ) = (20) k e 20 /k ! k = 0 , 1 , . . . otherwise (5) Problem 6.6.2 Solution Knowing that the probability that voice call occurs is 0.8 and the probability that a data call occurs is 0.2 we can define the random variable D i as the number of data calls in a single telephone call. It is obvious that for any i there are only two possible values for D i , namely 0 and 1. Furthermore for all i the D i ’s are independent and identically distributed withe the following PMF. P D ( d ) = . 8 d = 0 . 2 d = 1 otherwise (1) From the above we can determine that E [ D ] = 0 . 2 Var [ D ] = 0 . 2 . 04 = 0 . 16 (2) With these facts, we can answer the questions posed by the problem. (a) E [ K 100 ] = 100 E [ D ] = 20 (b) Var[ K 100 ] = p 100 Var[ D ] = √ 16 = 4 (c) P [ K 100 ≥ 18] = 1 Φ ( 18 20 4 ) = 1 Φ( 1 / 2) = Φ(1 / 2) = 0 . 6915 (d) P [16 ≤ K 100 ≤ 24] = Φ( 24 20 4 ) Φ( 16 20 4 ) = Φ(1) Φ( 1) = 2Φ(1) 1 = 0 . 6826 Problem 6.7.2 Solution (a) Since the number of requests...
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 Fall '08
 Duman
 Central Limit Theorem, Variance, Probability theory

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